\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2 +H_2\\ n_{H_2}=n_{FeCl_2}=n_{Fe}=0,15\left(mol\right)\\ a,V_{H_2\left(\text{đ}ktc\right)}=0,15.22,4=3,36\left(l\right)\\ b,m_{FeCl_2}=127.0,15=19,05\left(mol\right)\\ m_{\text{dd}FeCl_2}=m_{Fe}+m_{\text{dd}HCl}-m_{H_2}=8,4+200-0,15.2=208,1\left(g\right)\\ C\%_{\text{dd}FeCl_2}=\dfrac{19,05}{208,1}.100\approx9,154\%\)
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